MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $R$ is a local ring and let $I\subset R$ be some nontrivial ideal. Are there conditions that we can place on $I$ so that if $R/I$ is regular, then so is $R$?

I am aware of the result that states: if $R$ is already a regular local ring, then $R/I$ is regular iff $I$ is generated by a subset of a regular system of parameters. But I am wondering more about whether or not the regularity of $R$ itself can be determined by it's quotient $R/I$.

Background for this question: after reading the responses to the question When is a blow-up non-singular? I am trying to work through the first argument in the second section of the paper "On the smoothness of blow-ups." I think the author uses the result that I am asking about when they state: "$S_P/f_i S_P$ is a regular local ring; since $f_i$ is a non-zero-divisor on $S_P$, it follows that $S_P$ is itself regular." I am just trying to figure out where this result comes from.

share|cite|improve this question
up vote 2 down vote accepted

The quoted result relies on the following elementary characterization of local regular rings:

Let $R$ be a local ring with maximal ideal $\mathfrak m$ and $x\in\mathfrak m-\mathfrak m^2$. Then $R$ is regular iff $R/(x)$ is regular and $x$ doesn't belong to any minimal prime.

share|cite|improve this answer
Wow, that was a fast response to this question. I guess this is a result that is very well known, that I must have just overlooked in my studies. Thank you for the answer, and for the generalization and counterexample. – DavidWayne Jul 7 '13 at 19:12
@QiL'8 Maybe I'm missing something: in your case, if, for instance, $r=1$ and $R/(x)$ is regular, then the embedding dimension of $R/(x)$ equals $\dim R/(x)=\dim R-1$. I think now it follows that $x\notin\mathfrak m^2$. (What I want to say is that in the end the elements $x_i$ turn out to be outside of $\mathfrak m^2$, although you are not assuming this.) – user26857 Jul 7 '13 at 19:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.